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  4. npr=3024 and nCr=126 ,then r is:

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Q. $ ^{n}{{p}_{r}}=3024 $ and $ ^{n}{{C}_{r}}=126 $ ,then $ r $ is:

KEAMKEAM 2004

Solution:

Given, $ ^{n}{{P}_{r}}\,=3024 $
$ \therefore $ $ \frac{n!}{(n-r)!}=3024 $ and
$ ^{n}{{C}_{r}}=\frac{n!}{r!(n-r)!} $
$ \Rightarrow $ $ 126=\frac{3024}{r!} $
$ \Rightarrow $ $ r!=24=4! $
$ \Rightarrow $ $ r=4 $